# Particle in 3D box

This is the question:

For a particle in a three-dimensional box of sides a, b, and c, where a does not equal b and b=c, make a table of n_x, n_y, and n_z, the energies, and the degeneracies of the levels in which the quantum numbers range from 0 to 4 (Take ((a^2)/(b^2)) = 2).

Ok, I think I have an idea of what I'm supposed to do, but I'm a little confused on two parts of the question. When they ask for a table using quantum numbers from 0 to 4, would that mean I'd have to make a long list with n_x, n_y, and n_z like:
0 0 0
0 0 1
0 0 2
0 0 3
0 0 4
0 0 5

and so on until I've listed all 125? Someone in my class said that there were 64 because of the zeroes, however I didn't really understand why we would be able to. I know that:
http://img165.imageshack.us/img165/1706/01mk3.jpg [Broken]
So once I list all the states, I would have to substitute in the values for n_x, n_y, and n_z - but am I supposed to be able to get numerical values for the energies? I can get solve for a in terms of b and such, then plug in, but I still wouldn't get any numbers. Once I can figure out what I have to list, all I'd have to do to list the degeneracy is count the amount of states with different quantum numbers that have the same energy.

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I assume they mean $$n_x=1,2,3,4 n_y=1,2,3,4 n_z=1,2,3,4$$

Since b=c you get 40 different cases, not 64, because for instance $$n_y=2,n_z=1$$ and $$n_y=1,n_z=2$$ are energy equivalent cases. You can combine $$n_y,n_z$$ in 10 different non degenerate ways(not 16) and then combine that with the 4 possible cases for $$n_x$$

n can not be zero thats probably what your friend mean. Consider a particle in a one dimensional box. If you put n=0 it basicly means there is no particle. the lowest state is when all three n values are =1.

Numerical answeres doesnt seem to be possible but you can easily express them as multiples since $$\frac{a^2}{b^2}=2$$

Hope this helped a bit?

Azael said:
I assume they mean $$n_x=1,2,3,4 n_y=1,2,3,4 n_z=1,2,3,4$$

Since b=c you get 40 different cases, not 64, because for instance $$n_y=2,n_z=1$$ and $$n_y=1,n_z=2$$ are energy equivalent cases. You can combine $$n_y,n_z$$ in 10 different non degenerate ways(not 16) and then combine that with the 4 possible cases for $$n_x$$

I see what you mean now, although now I feel bad since I went ahead and listed 64 cases. At least now I can shorten my answer. Thanks a lot, this cleared up the bit about the zero.